Question

Let $D$ and $E$ be the midpoints of the sides $AC$ and $BC$ of a triangle $ABC$ respectively. If $O$ is an interior point of the triangle $ABC$ such that $OA +2 OB + 3 OC = 0$. then the area (in sq. units) of the triangle $ODE$ is

• A. 6
• B. 5
• C. $\frac{3}{4}$
• D. 0

Answer 1

Answer: (D)

Given,
$OA + 2OB + 3OC = O$

$\Rightarrow a +2 b +3 c =0 \dots$(i)
Now, area of $\Delta O D E$
$=\frac{1}{2}| O D \times O E |=\frac{1}{2}\left|\left(\frac{ a + c }{2}\right) \times\left(\frac{ b + c }{2}\right)\right| $
$=\frac{1}{8}| a \times b + a \times c + c \times b + c \times c | $
$=\frac{1}{8}| a \times b + a \times c + c \times b | \dots$(ii)
Multiply Eq. (i) by (b),
$a \times b +2( b \times b )+3( c \times b )=0$
$\Rightarrow a \times b +3( c \times b )=0 \dots$(iii)
Again multiply Eq. (i) by (c),
$a \times c +2( b \times c )+3( c \times c )=0$
$\Rightarrow a \times c +2( b \times c )=0 \ldots$ (iv)
Adding Eqs. (iii) and (iv), we get
$a \times b +3( c \times b )+ a \times c +2( b \times c )=0 $
$\Rightarrow a \times b +3( c \times b )+ a \times c -2( c \times b )=0 $
$\Rightarrow a \times b + a \times c + c \times b =0 \ldots$ (v)
$\therefore $ Area of $\Delta O D E=0 $ [By Eq. (v)]